# What's the problem with Black Hole evaporation?

Black hole evaporation is not unitary because it takes a pure state to a mixed state. On the other hand, ordinary decay processes in Quantum Mechanics do not seem very unitary either. (For example, if I know my final state is a Breit-Wigner distribution, I don't think it's possible in general to reconstruct the initial state.) In what way are decays any more unitary than BH evaporation?

• "Black hole evaporation is not unitary because it takes a pure state to a mixed state" - this statement is almost definitely false. The truth will become clearer as we gain better understanding of quantum gravity, but I think we can already be pretty confident it's false. For example, Hawking conceded en.wikipedia.org/wiki/… because he decided "black holes eventually transmit, in a garbled form, information about all matter they swallow" – Steve Byrnes Dec 19 '18 at 11:48
• That's a good point. I mean in classical gravity. – Eric David Kramer Dec 20 '18 at 12:18

On the other hand, ordinary decay processes in Quantum Mechanics do not seem very unitary either.

No, decay processes are perfectly unitary. However, they entangle the decaying object and the environment, so you can't recover the state by measuring the object or environment alone.

As a simple example, consider an atom with ground and excited states $$|g\rangle$$ and $$|e \rangle$$, inside a cavity which can have $$|0 \rangle$$ or $$|1 \rangle$$ photons in it, at frequency $$\omega = (E_e - E_g)/\hbar$$. The part of the Hamiltonian responsible for decay is $$H_{\text{int}} = \alpha |1g \rangle \langle 0e | + \text{h.c.}$$ and the evolution is perfectly unitary. For example, starting from the state $$|e0 \rangle$$ one smoothly evolves into a superposition of $$|e0 \rangle$$ and $$|g1 \rangle$$. The description of other quantum decay processes is similar, except that the environment will have more degrees of freedom. The state is pure at all times.

In what way are decays any more unitary than BH evaporation?

In the standard semiclassical derivation of Hawking radiation (which does not look at all like finding $$H_{\text{int}}$$ above) one finds that the emission spectrum is exactly thermal. So when the black hole completely evaporates, you're left with only thermal degrees of freedom, which correspond to a mixed state. That's the puzzle.

• I agree that the atom decaying in the cavity is a unitary process and is even reversible (journals.aps.org/prl/abstract/10.1103/PhysRevLett.58.353). However, if you take the cavity to have infinite volume, the decay will be irreversible even in the absence of decoherence. Moreover, if you wait an infinite amount of time, it will not be possible to recover the initial state (I think). – Eric David Kramer Dec 19 '18 at 12:52
• @EricDavidKramer Adding infinite volume and infinite space seems to me to simply muddy the issue. The black hole information paradox still exists even if the black hole is in a finite box. – knzhou Dec 19 '18 at 13:21
• @EricDavidKramer Furthermore even in infinite volume, the time evolution is still unitary. It may be hard in practice to recover the original state if the environment is big, but that's not the same as being impossible, and it's not the issue here. – knzhou Dec 19 '18 at 13:49
• I'm worried that the thermal spectrum of radiation might be correct only if you wait an infinite amount of time for the black hole to actually collapse. In that case, this would be similar to the limit of infinite volume and decay after an infinite time. Your point about the black hole in a finite box is a good one, and I'm unsure how to think about that. – Eric David Kramer Dec 19 '18 at 14:11
• @EricDavidKramer Perhaps, but now this is getting into a full discussion of the black hole information paradox. It goes without saying that I don't know the answer! Nor can I really say anything more for sure. – knzhou Dec 19 '18 at 14:47

In QFT with classical gravity, the problem with Black Hole evaporation is that the asymptotic spectrum is that of a thermal density matrix with respect to the free theory of an asymptotic observer. The asymptotic density matrix does not merely have Planck-spectrum expectation values - it is also completely diagonal, meaning that the coherence between the off-diagonal matrix elements vanishes.

This is because the particles escaping to infinity are entangled with the particles falling in to the black hole, and since there is no way to measure what fell in to the black hole, the reduced density matrix on the Hilbert space of particles escaping to infinity is that of a mixed state.

On the other hand, in a decay, the evolution is unitary for any finite time. The state will be the Breit-Wigner, plus some exponentially decaying transients. At any finite time, the exponentially decaying deviation from Breit-Wigner can (in principal) be measured exactly to reconstruct the original state.